#OK to post homework
#Joseph Koutsoutis, 04-14-2024, Assignment 22


#1 done

#2 I don't know physics well but here is my rough understanding of the argument:

# By the wikipedia article, we have that tidalStrength(moon) is proportional to mass(moon) / distFromEarth(moon)^3.
# Similarly, we have that tidalStrength(sun) is proportional to mass(sun) / distFromEarth(sun)^3.
# It is known that tidalStrength(moon) >> tidalStrength(moon), so 
# mass(moon) / distFromEarth(moon)^3 >> mass(sun) / distFromEarth(sun)^3.
# We next note that the volume of a sphere is proportional to its radius^3.
# We also note that since the eclipse occurs, we have that 
# radius(sun) / distFromEarth(sun) approximately equals radius(moon) / distFromEarth(moon).
# Therefore:
#     mass(moon) / distFromEarth(moon)^3 >> mass(sun) / distFromEarth(sun)^3 implies
#     mass(moon) / radius(moon)^3 >> mass(sun) / radius(sun)^3 implying
#     mass(moon) / volume(moon) >> mass(sun) / volume(sun) implying
#     density(moon) >> density(sun).

#3

# Currently I'm still generating the weight enumerators of BCH codes by brute force. I found that it wasn't 
# feasible to generate certain codes where the distance d was close to n/2, so I've only generated weight 
# enumerators for codes with <= 10^8 code words for all 3 <= 2t+1 <= n <= q-1 for all primes q <= 61 so far.
# To get more data, I've also started implementing the more general definition of BCH codes found on wikipedia
# (https://en.wikipedia.org/wiki/BCH_code), and I'll probably try to generate a lot of weight enumerators for codes
# with <= 10^8 words for primes/prime powers <= some Q. The OEIS has some weight distributions for some very large
# BCH codes (https://oeis.org/wiki/List_of_weight_distributions), and I'm planning on reading the articles linked
# to learn about the methods used to generate weight enumerators under certain nice conditions.

# I haven't had the chance to look at the data too closely yet, so this observation might be completely wrong but
# I noticed the following pattern for weight enumerators for the BCH codes as described by Hill. I'll only
# give 2 examples since I don't know a good way to describe it:

# Based on the output of my code (assuming its correct), here are the weight enumerators for a couple BCH codes with 
# fixed d=5,p=53 and increasing n (I've omitted larger n for clarity):

#d=5, n=5, p=53, enumerator=                                                    52*t^5+1
#d=5, n=6, p=53, enumerator=                                         2496*t^6+ 312*t^5+1
#d=5, n=7, p=53, enumerator=                            130312*t^7+ 17472*t^6+1092*t^5+1
#d=5, n=8, p=53, enumerator=               6775184*t^8+1042496*t^7+ 69888*t^6+2912*t^5+1
#d=5, n=9, p=53, enumerator=352311388*t^9+60976656*t^8+4691232*t^7+209664*t^6+6552*t^5+1

# If we want to compute the coefficient of t^5, I found that the terms in the sequence given by generating function
# 52/(1-x)^6 give them (Maple outputs series(52/(1-x)^6,x,5) = 52 + 312*x + 1092*x^2 + 2912*x^3 + 6552*x^4 + O(x^5)).

# Similarly, to compute the coefficients of t^6, we note that the first nonzero coefficient is 2496 (or 53^2 - 312), and
# we find that Maple outputs series(2496/(1-x)^7,x,4) = 2496 + 17472*x + 69888*x^2 + 209664*x^3 + O(x^4).

# More generally, it appears that if we want the coefficients of t^k, we just need the first term and can look
# at series(firstTerm / (1-x)^(k+1)) to find more coefficients.

# Again, I didn't do much testing but here's a second example to show it held for another output with different d,p:

#d=39, n=39, p=61, enumerator=                                           60*t^39+1
#d=39, n=40, p=61, enumerator=                             1320*t^40+  2400*t^39+1
#d=39, n=41, p=61, enumerator=              123660*t^41+  54120*t^40+ 49200*t^39+1
#d=39, n=42, p=61, enumerator=6826800*t^42+5193720*t^41+1136520*t^40+688800*t^39+1

# Maple outputs:

#  series(60/(1-x)^40,x,4) = 60 + 2400*x + 49200*x^2 + 688800*x^3 + O(x^4)
#  series(1320/(1-x)^41,x,3) = 1320 + 54120*x + 1136520*x^2 + O(x^3)
#  series(123660/(1-x)^42,x,2) = 123660 + 5193720*x + O(x^2)


#4 I didn't attempt this challenge.

#5 I didn't attempt this challenge.